Adobe Interview Questions

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

Rachita D. - "Start by explaining your understanding of Google Maps. It is a free web/mobile-based feature that allows users to navigate from Point A to B, trip planning locate restaurants, hair salons, other points of interest around a point on the map Businesses use to sponsor advertisements based on Map search string. Clarify what does improvement means, "Improve" is a loaded term - what do we mean by improvement: increased ad revenue, improve the quality of the results, increase downloads, imp"See full answer

Ashley C. - "I follow a variation of the RICE framework when prioritizing how I ship product features. I start by looking at: Reach: Because the customer segmentation across our product portfolio is so similar, I tend to hold a lot of weight on product features that will maximize our customer reach with a minimal LOE. Impact: After establishing which customer segments will benefit from the product feature, I determine the urgency and estimated impact on each customer segment based on customer i"See full answer

R S. - "One of the biggest risks I have taken was in a recent product, where we developed a new feature and released the feature externally without QA sign-off. Feature - Interactive Travel Itinerary Maps We had about ten days from the time we decided to include this new feature in the external launch. Usually, the biggest chunk of time in each release cycle was the sign-off from an external QA team. We fast-tracked the dev effort and completed the development of the feature in five days. Unfortunately,"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Venkata rakesh M. - "Here if we breakdown each dependency [A,B] , We need to check if there a cycle in Dependency Graph. If there is cycle installation is not possible, If there is no cycle installation is possible. Steps : 1: Build the graph 2: Perform DFS based Cycle Detection 3: Check each package if those packages have cycle or not."See full answer

Eshaa D. - "I started by reframing Adobe’s mission in my own words: to build creative and productivity tools that make everyday customer workflows easier. Then, I defined the goal: improving retention for the Scan app, as users often drop off to competitors such as Genius Scan or the native iOS Scanner after initial use. To better understand user behavior, I segmented users into three groups: Power users: Professionals such as lawyers or students who scan frequently. Low-engagement users: Inactive for 9"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

Anonymous Muskox - "Interviewer : How would you reduce ride cancellation on Uber ? Shailesh : Okay before I jump onto the solution, I would like to clarify some points for better understanding. So ride can be cancelled from the both the parties rider and the driver ? So which side are we talking about ? Interviewer : we are seeing more cancellation from diver side. Shailesh : From how long we are seeing this trend ? Interviewer : you can say from an year. Shailesh : Okay, also are we seeing these cancellations from"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Rahul M. - "Idea for solution: Reverse the complete char array Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters. vector reverseSubarray(vector& arr, int s, int e) { while (s reverseWords(vector& arr ) { int n = arr.size(); reverse(arr, 0, n - 1"See full answer

Aymen D. - "It would have been more interesting to focus on the system design rather than the Trie DS, Interviewee could have just mentioned the Trie and passed to things more important. Interviewee should have focused on the factors on which he wants to scale the API servers, popularity of the query parts ? region may be ? A hash of many factors ? Caches should have definitely be discussed, Cache eviction policies, Cache invalidation managements... Interviewee should have mentioned which kind of API pro"See full answer

Ajay U. - "// C++ program to print the count of // subsets with sum equal to the given value X #include using namespace std; // Recursive function to return the count // of subsets with sum equal to the given value int subsetSum(int arr[], int n, int i, int sum, int count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0)"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer

Victor H. - "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"See full answer